Testing Polynomial Equivalence by Scaling Matrices Markus Bl aser - - PowerPoint PPT Presentation

Testing Polynomial Equivalence by Scaling Matrices

Markus Bl¨ aser

Saarland University

Joint work with Raghavendra Rao and Jayalal Sarma (IIT Madras)

Polynomial equivalence testing

Input: f(X), g(X) ∈ K[x1, . . . , xn] (by blackbox access) Question: Is there an invertible matrix A such that f(X) = g(AX)? Polynomial identity testing:

◮ Given f ∈ K[x1, . . . , xn], is f identically zero? ◮ If f is given as a blackbox, then randomisation is inherently

needed for achieving polynomial running time.

◮ If f is given as a circuit, then it is a major open problem,

whether there is a deterministic polynomial time algorithm. Polynomial identity testing is a special case of equivalence testing. − → look for randomized algorithms

Previous results

Hardness:

◮ Ring isomorphism reduces to Equivalence testing

(Agrawal–Saxena ’06) Randomized polynomial time algorithms when g is a fixed polynomial:

◮ elementary symmetric polynomial (Kayal ’11) ◮ power sum (Kayal ’11) ◮ permanent (Kayal ’12) ◮ determinant (Kayal ’12) ◮ iterated matrix multiplication (Kayal et al. ’17)

Our results

We consider the case when A is a diagonal matrix.

◮ Randomized polynomial time algorithm for this case. ◮ Randomized polynomial time algorithm that gives a maximum

set of monomials such that their degree vectors are linearly independent. Kayal’s algorithms follow a general pattern:

◮ Reduction to permutation and scaling equivalence by studying

the corresponding Lie algebra.

◮ Random polynomials have a trivial Lie algebra. ◮ Reduction to scaling equivalence.

Related questions

What if A is not invertible?

◮ Kayal’12: NP-hard ◮ Using a result by Shitov, this can be strengthened to complete

for the existential theory over the underlying field.

◮ The question whether Xp(n)−n 1,1

pern(X) = detp(n)(AX) is equivalent to VP = VNP.

Isolating a monomial

Lemma (Mulmuley–Vazirani–Vazirani)

Let F1, . . . , Fm ⊆ {1, . . . , n}. If w1, . . . , wn ∈ {1, . . . , 2n} are chosen uniformly at random, then there is a unique set of minimum weight with probability ≥ 1/2. (w(F) =

i∈F wi) ◮ f ∈ K[x1, . . . , xn] multilinear ◮ monomials α · xe1 1 · · · xen n with ei ∈ {0, 1} ◮ xi → ywi ◮ unique monomial of minimum degree

Isolating a monomial (2)

Theorem (Klivans and Spielman)

There is a randomized algorithm that given a non-zero f ∈ K[x1, . . . , xn] (by blackbox access) outputs a monomial m of f (degree vector Deg(m) and coefficient) with probability ≥ 1 − ǫ in time polynomial in n, ∆, and 1/ǫ.

◮ Weighted sets −

→ larger weights.

◮ Since we only have blackbox access, substitutions are

simulated when evaluating. Instead of substituting xi → ywi and then evaluating at α, we directly evaluate f(αw1, . . . , αwn).

Extracting a degree basis

◮ monomial m = α · xd1 1 · · · xdn n ◮ degree vector Deg(m) := (d1, . . . , dn)

Definition (Degree basis)

Let f ∈ K[x1, . . . , xn]. Monomials m1, . . . , mt are a degree basis of f if for any monomial m of f, Deg(m) ∈ linR{Deg(m1), . . . , Deg(mt)}.

Theorem

There is a randomized algorithm, that given f ∈ K[x1, . . . , xn] (by black box access) outputs a degree basis of f. The running time is polynomial in the degree ∆, n, and the bit size L of the coefficients.

Extracting a degree basis (2)

◮ Let m1, . . . , mt monomials of f with ◮ linearly independent degree vectors v1, . . . , vt, t < n. ◮ Extend v1, . . . , vt to an (unknown) degree basis v1, . . . , v^ t. ◮ Let p be a prime such that v1, . . . , v^ t stay linearly

independent over Fp

◮ By the Hadamard bound and the prime number theorem, p is

small.

Extracting a degree basis (3)

◮ Let u1, . . . , un−t be linearly such that viuj = 0 over Fp for all

1 ≤ i ≤ t, 1 ≤ j ≤ n − t.

◮ If w is a vector not contained in lin{v1, . . . , vt}, then there is a

j such that wuj = 0 over Fp.

◮ Substitute xi → yuj,ixi, 1 ≤ i ≤ n, where uj,i are the entries

• f uj. Let fj be the resulting polynomial.

◮ This maps every monomial m to ydm for some d.

Lemma

• 1. The degree of fj is bounded by O(∆npolylog(∆n)) for all j.
• 2. If Deg(m) ∈ lin{v1, . . . , vt}, then for every j, p|d.
• 3. If Deg(m) /

∈ lin{v1, . . . , vt}, then there is a j such that p |d.

Extracting a degree basis (4)

Let fj =

∆j

• d=0

gd · yd. Recall: m → ydm.

◮ View fj as a polynomial in y with coefficients from

K[x1, . . . , xn].

◮ Extract a monomial from the coefficient polynomial of a

power yd with p |d using Klivans–Spielman.

◮ If we find a monomial, then we set vt+1 to be its degree

vector.

◮ If we do not find such a monomial, then v1, . . . , vt is a degree

basis.

Simulating blackbox access to gd

fj =

∆j

• d=0

gd · yd We have to provide blackbox access to the gd’s:

◮ Given blackbox access to f, it is easy to simulate blackbox

access to fj.

◮ Now assume we want to evaluate gd at a point ξ ∈ Kn. ◮ We evaluate fj at the points (ξ, αi) ∈ Kn+1, 0 ≤ i ≤ ∆j,

where the αi are pairwise distinct, that is, we compute values fj(ξ, αi) = ∆j

d=0 gd(ξ)αd i . ◮ From these values, we interpolate the coefficients of fj, viewed

as a univariate polynomial in y. The coefficient of yd is gd(ξ).

Polynomial equivalence

◮ Let f(X), g(X) ∈ K[x1, . . . , xn] (given by black box access). ◮ Degree of f and g is bounded by ∆ and all coefficients of f

and g have bit length ≤ L.

◮ Assume there is an invertible diagonal matrix A such that

f(X) = g(AX).

Observation

If f(X) = g(AX), then f and g have the same set of monomials. xd1

1 · · · xdn n → ad1 1 · · · adn n · xd1 1 · · · xdn n

Polynomial equivalence (2)

Lemma

Let S = {(mi, αi) | 1 ≤ i ≤ n} be a degree base of f. If f(X) = g(AX) for a non-singular diagonal matrix A, then such an A can be computed deterministically in time polynomial in n, ∆ and L. (ai given by polynomial size expressions with roots.)

◮ Let αi = 0 and βi = 0 be the coefficient of mi in f and g. ◮ We get n equations

αi = βi

n

• j=1

adi,j

j

where vi = Deg(mi) =: (di,1, . . . , di,n).

◮ Unique solution:

ai =

n

• j=1

(αj/βj)

¯ di,j

where D = (di,j) and D−1 = (¯ di,j).

Polynomial equivalence (3)

What if a degree basis of f has size t < n?

Lemma

Let a1, . . . , an be any solution to log αi = log βi +

n

• j=1

di,j log aj, 1 ≤ i ≤ t, and let A be the corresponding diagonal matrix. Let r(x) be a monomial with coefficient δ and degree vector u = (e1, . . . , en) contained in the linear span of v1, . . . , vt, i.e., u = λ1v1 + · · · + λtvt. Then the coefficient of r(Ax) is δ · α1 β1 λ1 · · · αt βt λt , in particular, it is independent of the chosen solution for a1, . . . , an.

Algorithm

Scaling equivalence test Input: Black box access to polynomials f, g ∈ K[x1, . . . , xn] Output: Nonsingular diagonal matrix A with f(x) = g(Ax) if such an A exists

1: Apply Gen-Mon with polynomial f to get a set S. 2: Apply Gen-Mon to g using the same random bits as above to

get a set S′.

3: If S and S′ do not contain the same degree vectors, then

Reject.

4: Solve for the entries of A using Lemma 6. 5: Accept if and only if f(x) − g(Ax) is identically zero.

Theorem

The Algorithm returns correct the correct answer with high

• probability. It runs in time polynomial in ∆, n and L.

Discussion

◮ Can our result extended to permutation and scaling

equivalence?

◮ Are the other polynomials g for which equivalence testing can

be done in polynomial time?